I need to make sure I understand perfectly the high-Z state and its relation to voltage.I intend to explain what I have understood based on three example, is there something wrong ?

Let's consider the three following examples : enter image description here

  • If the LED is red, the point 1 will approximatively be at 1.3 V because the voltage drop across a RED LED is approximatively 2 V.

  • Point 4 is at high-Z so we do not know its voltage. Can we assume it is near 1.3 V, or can it really be anything ? If I measure it with a voltmeter I would get around 1.3 V, but it is because the voltmeter is injecting a small current resulting in measuring the LED's resistor value ?

  • Point 3 is at high-Z as well. But is point 2 considered at high-Z ? It is not really an unconnected wire as it is connected to a resistor. More generally, if we replace the resistor with any other passive component, would 2 be at high-Z or no ?

  • \$\begingroup\$ @DKNguyen connections 2 and 4 not Hi-Z: If you connect wire with node 2 to ground, there will flow a significant current through that wire \$\endgroup\$
    – Huisman
    Apr 28, 2020 at 13:30
  • \$\begingroup\$ @DKNguyen Whether wire with node 3 is Hi-Z depends on the resistance of the resistor (without reference designator....) \$\endgroup\$
    – Huisman
    Apr 28, 2020 at 13:32
  • \$\begingroup\$ @Huisman Maybe I misunderstood what the OP was asking. \$\endgroup\$
    – DKNguyen
    Apr 28, 2020 at 13:33

3 Answers 3


Think at an Hi-Z wire in this way: can I connect its lead to any (reasonable) point in the circuit without having current flowing in the wire?

If yes, then the other lead of the wire is connected to a Hi-Z point of the circuit. A Hi-Z point is one connected to a very high resistance so no current can flow.

In your schematic NO wire (neither of 1,2,3,4) is Hi-Z because if you connect one of them to GND, current starts flowing and the LED lights up. UPDATE: Wire number 3 can actually be Hi-Z, provided that the resistor attached to it has a very high ohmic value.

Hi-Z is used for example in microcontrollers when they start-up. Given that any I/O pin can be connected to anything, and at the start time the microcontroller does not even know whether a certain own I/O pin is (will be configured as) an input or an output, those I/O are Hi-Z: you can pull them low or high, you can attach them to an input of some other IC, but nothing happens, they seem to be connected nowhere and no current can flow.

Another example where Hi-Z is desirable is in scope probes: a scope probe should be Hi-Z (as much as possible) in order to not pollute (or worse) the signal you are trying to measure.

  • \$\begingroup\$ Wire with node 3 can be Hi-Z when the resistor above it is high ohmic. \$\endgroup\$
    – Huisman
    Apr 28, 2020 at 13:34
  • \$\begingroup\$ You should add "can I connect its lead to any (reasonable) point in the circuit without having [a significant] current flowing said lead? \$\endgroup\$
    – Huisman
    Apr 28, 2020 at 13:35
  • \$\begingroup\$ @Huisman corrected, is it better now? \$\endgroup\$ Apr 28, 2020 at 13:36
  • 1
    \$\begingroup\$ @Huisman I think the probe example is good and it has nothing to do with the I/O of the preceding paragraph. The last two paragraphs are examples of why one should want Hi-Z. \$\endgroup\$ Apr 28, 2020 at 14:31
  • 1
    \$\begingroup\$ @Leoman12 For the buffer, if it is Hi-Z, it is; but once connected to a load, that point is no more Hi-Z (current, if any, would/could flow in the load). For pin 4 in the OP schematic, it is not Hi-Z; it would be if it would be connected to nothing, as you say, or if connected to a Hi-Z point, or connected to something via a very big resistor. From another point of view: a point is Hi-Z if you can connect it to any part of the circuit and (virtually) no current flows. \$\endgroup\$ Apr 28, 2020 at 16:59

Hi-Z just means high impedant.

So, when connecting such a wire to any voltage with any frequency, there will flow hardly no current through that wire, because $$ \frac{V}{Z}=I \text{ }\text{ }\text{ or }\text{ }\text{ } \frac{V}{\text{high impedant value}}=\text{very small current} $$

In your applications, this only applies to wire 3 when the resistor above it is high ohmic (and so, high impendant).

Reading the wikipedia page about High Impedance, I see there is a distinction made between analog and digital electronics.

In digital circuits, a high impedance (also known as hi-Z, tri-stated, or floating) output is not being driven to any defined logic level by the output circuit.

So, in that case, it depends on your application how these logical levels are defined...

In analog circuits a high impedance node is one that does not have any low impedance paths to any other nodes in the frequency range being considered.

This is what I wrote in the first part of my answer.

In your schematic the load connected to the nodes 1,2,3 or 4 determined whether the LED starts to (seriously) conduct. So, it is load dependent.
When you want to explain digital Hi-Z, you'd better use the following schematic:


simulate this circuit – Schematic created using CircuitLab

Here, the output pin can be configured as output high (by driving M1), low (by driving M2) and Hi-Z (by neither driving M1 or M2).

  • \$\begingroup\$ Why can I not start an answer with Hi???? \$\endgroup\$
    – Huisman
    Apr 28, 2020 at 13:54
  • 2
    \$\begingroup\$ Looks like a false positive for this meta.stackexchange.com/questions/2950/… \$\endgroup\$
    – Justin
    Apr 28, 2020 at 14:01
  • \$\begingroup\$ @Justin Good catch, that seems to be the culprit indeed. May not be worth adding an exception for this small use-case network wide. \$\endgroup\$
    – Mast
    Apr 29, 2020 at 7:14

If I measure it with a voltmeter I would get around 1.3 V, but it is because the voltmeter is injecting a small current resulting in measuring the LED's resistor value ?

Kind of. The voltmeter is not "injecting" a current. A voltmeter behaves (almost) like a very high resistor, so putting the voltmeter in there effectively turns the circuit into your first example, and you see 1.3V. The voltmeter's impedance is enormous, so in almost every other case you wouldn't notice a change in circuit behaviour with the voltmeter attached. But here, because the only thing completing the circuit is the voltmeter's impedance, you do.

If you didn't have the voltmeter there, you have no current through the LED, and hence no voltage drop on the LED. So point 4 will be at 3.3V. Unfortunately you can't prove that with a voltmeter though! :)


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